Q1(d). LDA at -78°C: kinetic enolate of 2-phenylcyclohexanone
Give the structures of C and D and account for their formation.
[structure] → (LiN(iPr)₂, -78°C, THF) C → (MeI) D (major)
Answer in brief
C = kinetic lithium enolate formed by removal of the C-6 hydrogen; D = 2-methyl-6-phenylcyclohexan-1-one (major). The more substituted 2-methyl-2-phenyl ketone would need thermodynamic conditions.
Key points
- C. The C-2 hydrogen is more acidic (benzylic) but is tertiary and shielded by the phenyl ring. The bulky (iPr)₂N⁻ removes a less hindered C-6 hydrogen through a cyclic six-membered transition state with Li bound to the…
- At -78°C with no excess ketone there is no proton exchange, so C does not equilibrate to the more stable, conjugated (thermodynamic) enolate C1=C2–Ph.
- D. The enolate α-carbon (C-6) attacks CH₃I by SN2 (C-alkylation): D = 2-methyl-6-phenylcyclohexan-1-one (the new methyl and the phenyl on the two different α-carbons; a cis/trans mixture that favours the diequatorial isomer on…
- Under thermodynamic conditions (NaH or Et₃N/Me₃SiCl with heating, or excess ketone) the conjugated enolate forms instead and methylation gives 2-methyl-2-phenylcyclohexanone.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. 2-Phenylcyclohexanone has two sets of α-hydrogens: one tertiary, benzylic H on C-2 and two secondary H on C-6. LDA is a strong, bulky, non-nucleophilic base; at -78°C in THF deprotonation is fast and irreversible, so the enolate formed fastest (kinetic enolate) is the one obtained.
Scheme: LDA removes the more accessible C-6 hydrogen; the enolate carbon attacks MeI.
- C. The C-2 hydrogen is more acidic (benzylic) but is tertiary and shielded by the phenyl ring. The bulky (iPr)₂N⁻ removes a less hindered C-6 hydrogen through a cyclic six-membered transition state with Li bound to the carbonyl oxygen. C = lithium 6-phenylcyclohex-1-en-1-olate (the less substituted enolate).
- At -78°C with no excess ketone there is no proton exchange, so C does not equilibrate to the more stable, conjugated (thermodynamic) enolate C1=C2–Ph.
- D. The enolate α-carbon (C-6) attacks CH₃I by SN2 (C-alkylation): D = 2-methyl-6-phenylcyclohexan-1-one (the new methyl and the phenyl on the two different α-carbons; a cis/trans mixture that favours the diequatorial isomer on equilibration).
- Under thermodynamic conditions (NaH or Et₃N/Me₃SiCl with heating, or excess ketone) the conjugated enolate forms instead and methylation gives 2-methyl-2-phenylcyclohexanone.
Conclusion. C = kinetic lithium enolate formed by removal of the C-6 hydrogen; D = 2-methyl-6-phenylcyclohexan-1-one (major). The more substituted 2-methyl-2-phenyl ketone would need thermodynamic conditions.
Application. Kinetic (LDA, -78°C) versus thermodynamic enolate control decides the site of alkylation and aldol reactions in steroid and terpene syntheses.
Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

All parts of Question 1
- 1(a) Aromatic, anti-aromatic or non-aromatic10 marks
- 1(b) E2 of 2-bromo-2-methylpentane: Zaitsev and Hofmann10 marks
- 1(c) Diethylamine and 2-fluoro-1-acetylcyclohexene: addition–elimination10 marks
- 1(d) LDA at -78°C: kinetic enolate of 2-phenylcyclohexanone10 marks
- 1(e) Electrocyclic closure of (2E,4E)-hexa-2,4-diene: FMO10 marks